Quantitative Aptitude Prep Tips for MBA
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An alcohol solution is formed by mixing two solutions. Solution 1 had a concentration of 20 percent and a volume of 3L while solution 2 had a concentration of 40 percent with a volume of 1L. The resultant solution was named solution 3. Three litres of solution 4 which had an unknown concentration was mixed with solution 3. What must be concentration of solution 4 if that the new resultant solution has a concentration of 50 percent?
An alcohol solution is formed by mixing two solutions. Solution 1 had a concentration of 20 percent and a volume of 3L while solution 2 had a concentration of 40 percent with a volume of 1L. The resultant solution was named solution 3. Three litres of solution 4 which had an unknown concentration was mixed with solution 3. What must be concentration of solution 4 if that the new resultant solution has a concentration of 50 percent?
Sol1 | 20percent (0.6) | 3L |
Sol2 | 40percent (0.4) | 1L |
Sol3 | 1L | 4L |
Sol4 | XL | 3L |
Sol5 | 3.5L | 7L |
1 + x = 3.5
x= 2.5
Required percent y =
Upvote 
Two geometric means a, b are inserted between x and y such that a3 + b3 = n (ab) A, where A is the arithmetic mean of x and y. What is value of n?
Two geometric means a, b are inserted between x and y such that a3 + b3 = n (ab) A, where A is the arithmetic mean of x and y. What is value of n?
x, a, b, y are in GP
a = xr
b = xr2
y = xr3
r =
rewrite,
a = x2/3.y1/3
b = x1/3.y2/3
a × b = x × y
a3 + b3 = x2y + y2 x
= xy (x+y)
= xy (2A) ……………
Hence, n = 2
Upvote A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of equilateral of the equilateral triangle to the radius of circle.
A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of equilateral of the equilateral triangle to the radius of circle.
Let the side of = a
CD = a/2
Let EC = x
In GEC
GE/EC = tan60 =
GE = x
DE = a/2 – x = FG = HF
GE = HG
x = a – 2x
x =
GE = x =
r = side of square ÷ 2
r=
ratio = a/r
=
=
Upvote A certain sum of money invested at CI becomes Rs 3380 in 2 years & Rs 3515.20 in 3 years. What is the sum invested & what is rate % per annum?
A certain sum of money invested at CI becomes Rs 3380 in 2 years & Rs 3515.20 in 3 years. What is the sum invested & what is rate % per annum?
(d):3515.2 – 3380 = 135.2 = SI on Rs. 3380 for one year.
By SI formula, 135.2 = or R = 4 %
Now we have 3380 = P (1.04)2 or P = Rs. 3125
Upvote Duronto Express in motion, 66 metre long overtakes a passenger train 88 metres long travelling at 30 kmph in the opposite direction in 0.168 minute. Find speed of the Duronto Express.
Duronto Express in motion, 66 metre long overtakes a passenger train 88 metres long travelling at 30 kmph in the opposite direction in 0.168 minute. Find speed of the Duronto Express.
(c):Let speed of passenger train be x m/s.
x = 6.94 m/sec = 25 kmph.
Upvote A 10 floor building is build by 2 men, 3 women and 4 boys in 10 days. If a man, a women & a boy work in the ratio 5 : 4 : 2. Find the time that 6 men, 4 women and 7 boys take to build a 16 floor building.
A 10 floor building is build by 2 men, 3 women and 4 boys in 10 days. If a man, a women & a boy work in the ratio 5 : 4 : 2. Find the time that 6 men, 4 women and 7 boys take to build a 16 floor building.
(c):We have 4 men = 5 women 1 man = women, 2 women = 4 boys, 1 women = 2 boys women = 2 * boys = or 1 man = women = boys
Now 2M + 3W+ 4B = 2 * B + 3 * 2B + 4B = 15 B Boys
Or 15 Boys do the work in 10 days (10 hectares) 6M + 4W + 7B = (6 * + 4 * 2 + 7) B = 30
Boy
Upvote During the months of October Krishna went to a temple every day. He offered 4 flowers to her favourite deity. At the end of the month, he noticed that he did not offer the same combination of flowers on any two days during the month. What is the minimum number of varieties of flowers offered by Krishna during the month?
During the months of October Krishna went to a temple every day. He offered 4 flowers to her favourite deity. At the end of the month, he noticed that he did not offer the same combination of flowers on any two days during the month. What is the minimum number of varieties of flowers offered by Krishna during the month?
Let he offered 'n' different flowers
October has 31 days
nC4 ³ 31
Minimum value of n can take 7
Upvote In a quiz, six team with two member one man and one woman each participated. During dinner all the 12 members were seated around a round table. Find the probability that no man sat to the left of the woman from this team, provided that men and women were alternatively seated.
In a quiz, six team with two member one man and one woman each participated. During dinner all the 12 members were seated around a round table. Find the probability that no man sat to the left of the woman from this team, provided that men and women were alternatively seated.
Total numbers of ways = 5 × 6
Numbers of ways in which men can sit such that no men sat to the left of the women from his team i.e. Derangement of 6
D6 =
= 265
F (0) = 5 × 265
Probability =
Upvote (d) : 411 + 412 + 413 + 414 + 415
= 411 (1 + 41 + 42 + 43 + 44) = 411 * 341.
The factors of 341 are: 1, 11, 31 and 341. Thus, we can say that the values in each of the three options would divide the expression.
Upvote Sanjay wrote 25 consecutive even natural numbers on the board. The average of these numbers is A and largest number is X. After erasing one of the numbers, the average become B. He replaced the erased number with (X + 2) and the average become C. Find the difference between A and C if it is known that B is 59.5.
Sanjay wrote 25 consecutive even natural numbers on the board. The average of these numbers is A and largest number is X. After erasing one of the numbers, the average become B. He replaced the erased number with (X + 2) and the average become C. Find the difference between A and C if it is known that B is 59.5.
It is given average of 24 numbers = 59.5
Sum = 59.5 ×24 = 1428
Let the numbers erased be ‘m’
1428 + m =
1428 + m = 50n + 650
50n – m = 778
Least value ‘n’ can take 16
(i) for n = 16, m =22
Series 2 (17, 18… 41)
i.e. 34, 36, 38 ….82
from this series we cannot remove 22. So, this case is not valid.
(i
Upvote